Download Physics 241 Superconductivity Questions and Answers

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Physics 241 Superconductivity Questions and Answers
Q1: I would think that @ higher temperatures, when electrons can move easier due to more
energy, the material would better conduct electricity. However, this is not the case. How come?
Am I thinking about superconductivity incorrectly?
A: Because electrons are fermions, only electrons near the chemical potential (also called the
Fermi energy) respond to any applied electric field. The chemical potential for most metals is
around 5 eV, which has an equivalent to about 55,000 degrees Kelvin. For that reason, the
sample temperature change is negligible compared to the temperature equivalent of the electrons
at or near the Fermi energy. So the change in sample temperature plays essentially no role in
electrons moving faster or slower.
However, changing the sample temperature does play an important role- in changing the size of
the lattice vibrations. Lattices vibrate- the nuclei move- because of temperature. If the lattice
(nuclei) did not move at all, then the moving electrons would see every part of the lattice as the
same as every other part and would not have any scattering due to changes in electrical potential
between one part of the lattice and another. However, at finite temperatures the lattice does
vibrate, and these vibrations cause changes in the electrical potential the electrons “see” and thus
cause scattering of the electrons. This scattering of the electrons is the main source of electrical
resistance, and for this reason electrical resistance increases as the sample temperature increases.
None of the above has anything to do with superconductivity. If the material is in the
superconducting state, electrical current is carried by Cooper pairs of electrons that essentially do
not scatter off the nuclei. For that reason, the electrical current carried by a superconducting
material- called the supercurrent- has virtually no electrical resistance.
Q2: How do Cooper pairs interact with the lattice to cause superconductivity?
A: In what are called low-Tc superconductors, essentially those discovered prior to the cuprates
in 1986, two electrons form a Cooper pair indirectly. One electron electrically attracts nearby
nuclei, which lowers the Coulomb potential energy. The second electron of the pair encounters
this lower potential energy and for that reason a pair of electrons has a lower total energy that the
two separate electrons. If we think of the two electrons forming a Cooper pair as doing so by
having a “glue”- exchanging a boson that causes their energy to lower- then the boson is a lattice
vibration, a phonon. Starting with the cuprates many superconductors have been discovered in
which the pairing that makes Cooper pairs involves magnetic excitations rather than lattice
vibrations, and the glue, the exchange boson, is a magnon (which is a unit of magnetic excitation
energy).
Q3: How do alloys work on the atomic level to increase TC ?
A: I do not know the full answer to this question. For low-Tc superconductors (before the
cuprates) the key material properties included the crystal structure (higher symmetry leads to
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higher Tc value) and more electrons with energy near the chemical potential (which leads to
higher Tc value). So the crystal structure and the electronic structure both play roles and affect
(Tc ). However, for the cuprates and the iron-based superconductors, the exact role that alloying
plays remains unclear.
Q4: What is a Cooper pair?
A: A Cooper pair is two (a pair) of electrons that attract each other and thereby make having the
Cooper pair lower in total energy than the two electrons separately. Since a pair of electrons have
an integral value of angular momentum, the Cooper pairs are bosons. As bosons, any number of
Cooper pairs can exist, which means macroscopic superconductivity can exist.
Q5: There is a phase change of current over a Josephson junction. What is there a phase change?
Why does the current have a phase?
A: Consider a wavefunction (Ψ) that represents the superconducting state. Suppose I change this
wavefunction to (Ψ' = Ψ eiΘ ). Then both wavefunctions would have the same probability
amplitude since │Ψ│2 is the same. If this was the whole story, then changing the phase of the
wavefunction would have no physical effect.
However, there is more to the story. In the presence of an electric and/or magnetic field, one of
Maxwell’s equations says:
•B=0
From this we can write:
B= x A
Here A is the electromagnetic vector potential. However, A is not unique. I can
write a different vector:
A1 = A - (f)
This new vector (A1 ) differs from A by the gradient of a scalar potential, but it
produces the same magnetic field.
When there is an electromagnetic potential included in the energy, then the
Schrödinger equation becomes:
(1/2m) (p + (e/c) A )2 Ψ = (E + eφ) Ψ
The equation remains unchanged if we transform the potentials by:
A → A - (f)
Φ →φ + (1/c) ∂f/∂t
f = (ħc/e) Θ
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With this change the wavefunction becomes:
Ψ (exp(-ief/ħc))
This is the phase change of the wavefunction. The probability current of this
wavefunction is:
J = (ħ/2mi) (Ψ* Ψ - ΨΨ* ) which changes if the phase of the wavefunction
changes. The Josephson effects are based on this phase of the wavefunction.
Q6: What is the relationship between spin and Boson/Fermion?
A: If the total angular momentum of an object is an integral multiple of (ħ), then the object is a
boson. If it is a half-integer multiple of (ħ), then it is a fermion.
Q7: If you change the lattice structure of a material using a method like “splat”, does this change
TC ?
A: Generally yes. If the symmetry of the crystal lattice (the arrangement of the nuclei) changes,
then the superconducting transition temperature (Tc ) usually changes as well.
Q8: How does the current through a superconductor affect the magnetic field?
A: The same current produces the same magnetic field whether it is a current through a metal or
a superconductor.
Q9: What is the reasoning for why superconductors can allow electrons to pass through a lattice
like a wave?
A: There may be a misunderstanding behind this question. Whether the material is a metal or in
the superconducting state, electrons move through the lattice as traveling waves. The material
does not have to be in the superconducting state for this to occur.
Q10: Why is superconductivity so important? What is the threshold that defines
superconductivity versus semiconductors?
A: To the extent it is important, superconductivity is important for some technological and basic
scientific reasons. The most important technological reason is having superconducting wires.
Because a supercurrent has no electrical resistance, and thus does not heat the wire and change
the electrical resistance, the supercurrent is much more stable that the electrical current through a
metal wire. Consequently, the magnetic field produced by the supercurrent is also more stable.
Having very stable magnetic fields allows much more precise measurements. These
measurements range from magnetic resonance imaging (MRI) pictures to precise magnetic
properties of materials.
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Another, but less important, technological reason is that superconductors are perfect diamagnets.
They expel any applied magnetic fields. This is used in building magnetically levitated trains,
which reduce friction and allow the trains to move at higher speeds. Another application of this
perfect diamagnet property is to use the superconductor as a magnetic shield.
As to the basic science, superconductors were the first types of materials discovered for which
their lowest energy state was a single, macroscopic, many-body quantum state. As such,
superconductors serve as the prototype of many-body effects in materials.
Regarding the second question, this is complicated. Until 1986, all materials that became
superconductors were metals. So the phase transition was metal ↔ superconductor. However,
the cuprates and the iron-based superconductors are qualitatively different. For the low carrier
density part of the phase diagram, the phase transition is between insulator ↔ superconductor,
while for the higher carrier density part of the phase diagram the phase transition is still between
metal ↔ superconductor.
Q11: I’ve heard that the superconductivity of a material is dependent on a superconductive gap,
call it (Δ). I’ve also heard that this gap can be considered as a wavefunction, the SC
wavefunction, which gives different solutions for different materials. My question is: what is
occurring in the gap? How does what is occurring in the gap with its representative
wavefunction? Another question is: what property of the material brings about the formation of
Cooper pairs at T = TC ? What precisely is a Cooper pair?
A: These questions are most interesting. I have tried to explain what a Cooper pair is in question
two (Q2). I confirm what you said: a material that is a superconductor has a superconducting gap
energy that is commonly denoted (Δ). This gap is an energy gap, which means that in an energy
range of (2Δ) centered about the chemical potential no single electron state exists. However, the
superconducting gap is not, repeat not, a wavefunction. It is an energy. On the other hand, the
size of the superconducting gap does indeed vary- change- with different materials.
As to your question on what is occurring in the gap? This is the most interesting question of all.
In the picture that Bardeen, Cooper and Schrieffer developed, the BCS theory, nothing happens.
One reason for this is that all the electrons at the chemical potential, regardless of what direction
they move in, experience the same size superconducting gap as the temperature is reduced below
the critical temperature (Tc ). However, for the cuprates and the iron-based superconductors, the
situation is more complicated. One reason for the complications is that the size of the
superconducting gap varies with the direction of the electron wavevector (k). Recall that in free
space the linear momentum of the electron (p) and the wavevector are related by:
p=ħk
In fact, for the cuprates there are wavevector values for which the superconducting gap (Δ) size
is zero at all temperatures for which we have data. This means that scattering and other
wavevector- dependent properties are more complicated for cuprates and iron-based
superconductors.
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Q12: What makes certain alloys more effective at superconductivity than others?
A: I do not know beyond what I said in response to question 3 (Q3).
Q13: Can you explain in more detail the properties of Cooper pairs?
A: I do not believe I can provide additional detail beyond what I have written above.
Q14: How do magnetic fields impact superconductors? I understand that superconductors are
diamagnetic but how does a superconductor function within a strong magnetic field?
A: There are two qualitatively different ways that different superconductors function when a
magnetic field is applied to the material. For what are called Type I (type one) superconductors,
the sample remains superconducting except for a small region called the penetration depth near
the surface of the material. This continues and continues as the magnetic field strength increases
until a critical field (Hc ) is reached, at which field the entire sample abruptly changes from a
superconductor to a metal and superconductivity is lost.
For all superconductors, there is a maximum current per area (current density J) that the material
can have and remain a superconductor. So even at very low temperatures and with no magnetic
field applied there is a critical current (Jc ) above which the entire material becomes a metal and
superconductivity is lost. However, (Jc ) decreases to zero both as the temperature increases
toward (Tc ) and as the applied magnetic field increases. In fact the critical field (Hc ) is
essentially zero for temperatures just below (Tc ) and increases to a maximum value at low
temperatures.
In addition to Type I superconductors, there are also the more technologically useful type II (type
two) superconductors. These are more technologically useful for the simple reason that the
maximum magnetic field for which they remain superconducting is typically much higher than
Type I superconductors.
In Type II superconductors, the materials responds to an applied magnetic field in a way
qualitatively different from Type I superconductors. At first, for very low fields, the response is
the same: the entire sample remains superconducting except for a small region near the surface.
When the applied magnetic field reaches what is called the lower critical field (Hc1 ), the
magnetic field begins to penetrate the material. However, most of the material remains
superconducting. Only very small regions, each small enough for a single magnetic flux quantum
(= h/2e), becomes normal metal material. This means that the supercurrent continues and the
electrical resistance remains zero. However, the maximum supercurrent (Jc ) decreases. This
pattern continues as the magnetic field gets larger. More and more of the material becomes a
normal metal, the supercurrent remains, electrical resistance remains zero, but (Jc ) continues to
decrease. At some magnetic field, called the upper critical field (Hc2 ) the entire sample is normal
metal, and there is no superconductivity remaining.
Q15: Confusion regarding Cooper pairs. What do Cooper pairs actually do? How are they
formed? Are they present in normal non-superconductors and if not why not?
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A: Cooper pairs carry an electrical current, the supercurrent, with no electrical resistance. They
are formed by some type of attractive interaction between two electrons. This can be lattice
vibrations (phonons) or magnetic excitations (magnons). To date we have not discovered any
material in which 3, 4, 5,… or any number larger than two of electrons are attracted to each other
in this fashion.
But, I congratulate you on your last question, which is very deep. In BCS superconductors,
Cooper pairs do not exist in the normal metal state. The reason is that there is no net attractive
interaction between pairs of electrons. The situation is almost exactly the same in a mean-field
ferromagnet. In both instances, once the long-range order (ferromagnetic or superconducting) is
gone, there is no short-range order on any distance scale.
However, there are experiments that establish the existence of short-range magnetic order for
some ferromagnets at temperature above the Curie temperature. As to whether there are Cooper
pairs above (Tc ) that is still a matter of debate. One of my friends and colleagues, Ivan Bozovic,
is a leader in investigating this question and reading his articles, although difficult, may be
rewarding.
Q16: Can you go over some homework questions or sample questions about superconductivity?
A. Sure Let us plan to do this November 11 or later after I return.
Q17: The difference between the two “types”: Type I (most metals) and Type II (most alloys).
Just the different properties of both and why it matters.
A: Please read my answer to question 14 (Q14).
Q18: I have a big problem in the second question of the midterm exam. I hope you can talk it in
detail and add some background info from it.
A: Sure. We can go over this once I return November 11.
Q19: Why are superconductors only formed by semiconductors? What happens to electrons in a
superconductor? Are all electrons in each atom liberated including non-valence electrons? Just
valence electrons?
A: There is a misconception here. Superconductors are formed only from metals. The only
electrons in a metal that participate in superconducting Cooper pairs are electrons near the
chemical potential in energy. So electrons that are in atomic energy levels of individual nuclei do
not participate in superconductivity.
Q20: What is the Fermi-Dirac distribution, Fermi gas and other things that came up on
homework? How is gap calculated?
A: The Fermi-Dirac distribution function is:
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fFD (E) = 1/ [exp(E- EF ) + 1]
Here (EF ) is the chemical potential, also called the Fermi energy. A Fermi gas is a collection of
objects that obey Fermi-Dirac statistics but have no potential energy interaction between them.
We will discuss this further later in our class.
Q21: How do physicists discover high-temperature superconductors? Where do they get this
folklore?
A: Much of the ‘discovery’ is luck. There is another large part, which is carefully looking for
commonalities in previous discoveries. This is the basis for various ‘rules’ of finding higher Tc
superconductors. Journal articles and conferences serve to transmit these common views.
Q22: Can superconductors be used to make extremely strong magnets? Or do they always avoid
magnetic fields?
A: Superconductors are adversely affected by external magnetic fields, and would not be used to
make magnets, extremely strong or otherwise. However, superconducting wires are used to wind
around magnetic pole pieces and to provide stable currents, which means stable magnetic fields.
Q23: What is the mechanism that prevents material from being superconducting all the time?
A: I do not know. Part of the story is that increasing the temperature means that the entropy
(disorder) term in the free energy, which lowers the total energy, becomes larger in magnitude.
This favors disorder, and works against any ordered phase, including superconductivity.
Q24: What exactly are the properties of a Cooper pair?
A: Please see my discussions above, especially my answer to question two (Q2).
Q25: What are Cooper pairs? How do they form? How do superconductors interact with magnets
and other materials? Does the bandgap of a superconducting material shrink as the temperature
of the material approaches the critical temperature?
A: As to what Cooper pairs are and how they form, please see my discussions above. Magnets
act on superconductors to disrupt the attractive interaction of Cooper pairs and thus destroy
superconductivity. I cannot say exactly how superconductors interact with other materials, since
their interactions vary with the type of material. Yes, indeed, the superconducting gap (Δ)
shrinks to zero as the material temperature approaches the critical temperature from below.
Q26: What are the main take home points that I need to get from this class? What is the
absolutely most important topics about superconductivity? It is hard to know what the things I
should be learning are since there is so much information being thrown our way. At the basic
need to know/understand level.
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A: The most important points to know about superconductors include:
*Superconductivity is a phase transition from a metallic state for low-Tc superconductors and
from an insulating state for low carrier concentration cuprates;
* Superconductivity is a second order phase transition and the specific heat diverges at the
critical temperature;
* All superconductors found to date form Cooper pairs in the superconducting state, have a nonzero bandgap (Δ) in the electronic energy levels, have a critical supercurrent and either one or
two critical magnetic fields, exhibit zero electrical resistance and act as perfect diamagnets,
expelling external magnetic fields.
Q27: I read the chapters. What is the Fermi-Dirac distribution and how do we use it to
understand superconductors? I know that once one Cooper pair forms others are more likely to,
but what causes that initial formation? Could you please describe what exactly the Fermi energy
is and how to use it in describing the band theory of conduction?
A: The Fermi-Dirac distribution function is:
fFD (E) = 1/ [exp(E- EF ) + 1]
We do not use it as such as understand superconductors. Your question about Cooper pairs
contains a misconception. You say that once a single Cooper pair forms others are more likely to.
In fact, for BCS superconductors the superconducting state is a single, macroscopic quantum
state having a certain density of Cooper pairs. The density of Cooper pairs at low temperatures is
approximately (10-2 – 10-4 ) of the carrier density at the Fermi energy. As the temperature
approaches (Tc ) from below, the density of Cooper pairs decreases to zero. However, forming
one Cooper pair does not make it more likely to form others.
The Fermi energy is the highest energy level occupied by electrons in a metal. The Fermi energy
does not by itself control electrical conduction in metals.
Q28: How does a bandgap and Fermi energy relate to superconductivity? The equation Eg = 3.5
kTc seems to indicate that a higher superconducting transition temperature causes a larger
bandgap to form but how does EFERMI relate to this?
A: Your question raises several interesting parts. Yes, a higher superconducting transition
temperature is accompanied by a larger maximum superconducting gap. The equation you cite is
for BCS superconductors, where the size of the gap is the same for all electrons at or near the
Fermi energy. This equation is not accurate for cuprate or iron-based superconductors. Since
superconductors arise from metals, any material with a semiconductor or insulator bandgap will
not be superconducting at any temperature. The value of the Fermi energy as such does not relate
to superconductivity. The number of electronic states per energy at the Fermi energy does relate
to superconductivity for BCS superconductors. Specifically, one of Matthias’s rules is that
materials with a higher density of electronic states at the Fermi energy are likely to have a higher
value of (Tc ). However, Matthias’s rules do not apply to the cuprates or the iron-based
superconductors.
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Q29: If BCS theory works for low-temperature superconductors why is it insufficient for these
high-temperature superconductors?
A: I do not know the full answer. Part of the answer is that the high-temperature superconductors
(cuprates, MgB2 , and iron-based) involve exchange bosons that are magnetic rather than phonon
(lattice vibration). Another part of the answer lies in the idea of a mean field theory. Any mean
field theory assumes that the volume corresponds to one object- in the case of superconductivity
one Cooper pair- also contains the effect of many other objects. For low-Tc BCS
superconductors, this definitely works. For example, the coherence length is a rough measure of
the ‘size’ of one Cooper pair. For any mean field theory to work, the volume corresponding to
the size of one Cooper pair should contain many other Cooper pairs so that the interaction can be
taken as the average of many pairs. In lead, the number of pairs in the volume of one Cooper pair
is about 104. In a typical cuprate, the number of pairs is about one. So a mean field theory such
as BCS is unlikely to work for the cuprates.
Q30: How can I use the equations of superconductivity to solve practical problems?
A: Honestly, anyone wanting to solve real engineering problems involving superconducting
materials will need to take at least one additional course that includes more details and more
example problems in the area of superconductivity. Because Physics 241 is a survey course, I
regret to say you will not be prepared to solve practical superconductivity problems at the end of
our course.